Question: The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $22.4$ years; the standard deviation is $1.5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living less than $19.4$ years.
Answer: $22.4$ $20.9$ $23.9$ $19.4$ $25.4$ $17.9$ $26.9$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $22.4$ years. We know the standard deviation is $1.5$ years, so one standard deviation below the mean is $20.9$ years and one standard deviation above the mean is $23.9$ years. Two standard deviations below the mean is $19.4$ years and two standard deviations above the mean is $25.4$ years. Three standard deviations below the mean is $17.9$ years and three standard deviations above the mean is $26.9$ years. We are interested in the probability of a tiger living less than $19.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the tigers will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the tigers will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $19.4$ years and the other half $({2.5\%})$ will live longer than $25.4$ years. The probability of a particular tiger living less than $19.4$ years is ${2.5\%}$.